First law of thermodynamics — dU = dQ − dW, sign conventions: Worked examples

1.7.13 · D3 · Physics › Thermodynamics › First law of thermodynamics — dU = dQ − dW, sign conventions

Kuch bhi karne se pehle, sirf teen quantities aur ek rule yaad karo jo hum is poore page par use karte hain.

Recall Woh ek rule (parent se)

$Δ U = Q - W$

$Q$ = gas mein daali gayi heat ( $+$ andar, $-$ bahar).
$W$ = gas dwara kiya gaya work ( $+$ jab expand kare, $-$ jab compress ho).
$Δ U$ = internal energy mein change ( $+$ = ideal gas ke liye hotter). Neeche sab kuch bas isme plug karna hai, dono signs ka dhyan rakhte hue.

Scenario matrix

First Law mein exactly do inputs hain jo sign carry karte hain: heat $Q$ aur work $W$ . Dono positive, negative, ya zero ho sakte hain. Yeh ek $3 \times 3$ grid hai — cases ki poori duniya. Neeche, har cell physical situation ka naam deti hai aur us example ki taraf point karti hai jo use nail karta hai.

	$W > 0$ (gas expands)	$W = 0$ (fixed volume)	$W < 0$ (gas compressed)
$Q > 0$ (heat in)	Ex 1 — heat in, work out	Ex 3 — isochoric heating	Ex 2 — heat in aur squeezed
$Q = 0$ (adiabatic)	Ex 5 — adiabatic expansion	Ex 6 — degenerate: free expansion	Ex 4 — adiabatic compression
$Q < 0$ (heat out)	Ex 7 — heat out phir bhi expands	Ex 3 (variant)	Ex 8 — dono signs negative

Do aur cells jo 3×3 grid mein fit nahi hoti par dikhani zaroori hain:

Special-shape work — isobaric ( $W = P Δ V$ directly), isothermal ( $Δ U = 0$ ). → Ex 9
Word problem / exam twist — ek cyclic process jahan $Δ U = 0$ ek poore loop mein. → Ex 10

Neeche ki figure wahi grid hai jo picture ke roop mein bani hai — diagonal (top-left se bottom-right) ko "boring cases jahan $Q$ aur $W$ $Δ U$ ko opposite ways mein push karte hain" ke roop mein memorise karo.

Figure — First law of thermodynamics — dU = dQ − dW, sign conventions

Examples

Cell $(Q > 0, W > 0)$ — heat in, gas bahar push karta hai

Cell $(Q > 0, W < 0)$ — heat in AUR squeezed bhi

Worked example Example 2 — Heat add ki jabki compress bhi ho rahi hai

Ek gas compress hoti hai (surroundings us par $150$ J kaam karte hain) aur saath mein $90$ J heat bhi absorb karti hai. $Δ U$ nikalo.

Forecast: Energy DONO doors se andar aa rahi hai (heat in, compression in). Toh $Δ U$ ek bada positive number hona chahiye — $90$ ya $150$ akele se bhi zyada.

Signs fix karo. Heat in: $Q = + 90$ . Gas par kiya gaya kaam $= 150$ J, toh gas dwara kiya gaya kaam $W = - 150$ hai. Yeh step kyun? Hamara $W$ hamesha gas dwara kiya gaya kaam hota hai; surroundings kaam karte hain toh sign flip ho jaata hai.
Plug in karo. $Δ U = Q - W = 90 - (- 150)$ . Yeh step kyun? Negative $W$ subtract karna exactly hai "compression energy andar pump karta hai."
Compute karo. $Δ U = 90 + 150 = + 240$ J.

Verify: Dono channels energy add karte hain: $90 + 150 = 240$ J store hua. Double-positive "dono doors khule" se match karta hai. ✓

Cell $(Q > 0, W = 0)$ — fixed volume (isochoric)

Worked example Example 3 — Isochoric heating (aur uska heat-out variant)

Ek rigid sealed container mein gas $80$ J heat absorb karti hai. $Δ U$ nikalo. Phir: agar woh $80$ J khoti hai, toh $Δ U$ nikalo.

Forecast: Rigid matlab walls hil nahi sakti, toh gas kuch bhi push nahi kar sakti. Kaam ke bina, saari heat internal energy ban jaati hai. Guess karo $Δ U = \pm 80$ J.

$W = 0$ kyun hai? Volume fixed hai ⇒ $d V = 0$ ⇒ $W = \int P d V = 0$ . Yeh step kyun? Kaam sirf volume change ke zariye hota hai; koi volume change nahi, koi kaam nahi — dekho Work done in thermodynamic processes (PV diagrams).
Heat-in case. $Δ U = Q - 0 = + 80$ J.
Heat-out variant. Ab $Q = - 80$ : $Δ U = - 80 - 0 = - 80$ J (gas thandi hoti hai).

Verify: $W = 0$ ke saath, $Δ U$ exactly $Q$ ko follow karta hai, sign aur sab. $+ 80$ garm karta hai, $- 80$ thanda karta hai. ✓ Isliye constant volume par heat capacity $C_{v}$ "sabse saaf" wali hai — dekho Heat capacities Cp and Cv.

Cell $(Q = 0, W < 0)$ — adiabatic compression

Worked example Example 4 — Adiabatic compression

Ek gas adiabatically compress hoti hai; us par $200$ J kaam kiya jaata hai. $Δ U$ aur temperature trend nikalo.

Forecast: Koi heat escape nahi kar sakti (adiabatic), phir bhi hum compress karke energy andar push karte hain. Uske jaane ki jagah sirf internal energy hai → gas garam hoti hai. Guess karo $Δ U = + 200$ J.

$Q = 0$ kyun? Adiabatic matlab "koi heat boundary cross nahi karti" — dekho Isothermal, adiabatic, isobaric, isochoric processes. Yeh step kyun? $Q = 0$ set karna adiabatic ki definition hai; yeh ek term khatam kar deta hai.
Work sign fix karo. Gas par kaam $= 200$ J ⇒ $W = - 200$ .
Plug in karo. $Δ U = 0 - (- 200) = + 200$ J ⇒ garam hoti hai.

Verify: Energy conservation sirf ek khule door ke saath (compression): saare $200$ J ruk jaate hain. "Diesel engine compress karne par garm ho jaata hai" wali intuition se match karta hai. ✓

Cell $(Q = 0, W > 0)$ — adiabatic expansion

Degenerate cell $(Q = 0, W = 0)$ — vacuum mein free expansion

Worked example Example 6 — Free expansion (dono terms zero)

Ek ideal gas evacuated chamber (vacuum) mein rush karti hai. Koi heat exchange nahi hoti aur woh kuch bhi push nahi karti. $Δ U$ aur $Δ T$ nikalo.

Forecast: Agar kuch bhi wapas push nahi kiya, toh koi kaam nahi hua — aur koi heat nahi bahi. Kuch bhi boundary cross nahi kiya. Toh $Δ U$ exactly $0$ hona chahiye.

$W = 0$ kyun? Gas vacuum mein expand karti hai — koi piston nahi, koi surroundings resist nahi karte. Force jo woh push karte hain $= 0$ , toh $W = 0$ chahe $d V > 0$ ho. Yeh step kyun? Yahi trap hai: volume sach mein badhta hai, par kaam ke liye kuch push karne ko chahiye. Zero opposing pressure ⇒ zero work.
$Q = 0$ kyun? Insulated / sudden ⇒ koi heat cross nahi karti.
Plug in karo. $Δ U = 0 - 0 = 0$ J.
Temperature. Ideal gas ke liye $U = U (T)$ sirf, toh $Δ U = 0 \Rightarrow Δ T = 0$ — dekho Internal energy and degrees of freedom.

Verify: Dono channels band ⇒ internal energy untouched ⇒ temperature unchanged. Yahi classic result hai: ideal free expansion gas ko thanda nahi karti. ✓

Cell $(Q < 0, W > 0)$ — surprising wala: heat OUT phir bhi gas EXPAND karti hai

Worked example Example 7 — Heat lose karte hue expand karna

Ek gas expand hoti hai, $300$ J kaam karti hai, lekin saath mein surroundings ko $100$ J heat khoti bhi hai. $Δ U$ nikalo.

Forecast: Energy DONO doors se bahar jaati hai (heat out, work out). Toh $Δ U$ strongly negative hona chahiye — gas bahut thandi hoti hai. Guess karo $- 300$ se neeche? ya exactly... chalte hain compute karte hain.

Signs fix karo. Heat bahar: $Q = - 100$ . Kaam gas dwara (expands): $W = + 300$ . Yeh step kyun? Dono internal energy par drain hain, lekin formula mein apne alag signs ke saath enter hote hain — dhyan zaroori hai.
Plug in karo. $Δ U = Q - W = - 100 - 300$ .
Compute karo. $Δ U = - 400$ J.

Verify: Bahar jaane wali total energy $= 100$ J (heat) $+ 300$ J (work) $= 400$ J bahar ⇒ $Δ U = - 400$ J. Dono-doors-out picture match karti hai. ✓

Cell $(Q < 0, W < 0)$ — dono negative

Special-shape work — pehle isobaric phir isothermal

Worked example Example 9 — Isobaric expansion, phir ek isothermal check

(a) Ek gas constant pressure $P = 2 \times 1 0^{5}$ Pa par $V_{1} = 1 \times 1 0^{- 3}$ m³ se $V_{2} = 3 \times 1 0^{- 3}$ m³ tak expand hoti hai, $Q = 700$ J absorb karti hai. $W$ aur $Δ U$ nikalo. (b) Alag se, ek ideal gas isothermally expand hoti hai $Q = 400$ J absorb karti hui. $W$ nikalo.

Forecast (a): Constant pressure matlab kaam sirf $P$ times swept volume hai — PV diagram par ek aasaan rectangle. Kuch sau Joules kaam ki umeed karo.

(a) Constant $P$ par kaam. $W = P Δ V = P (V_{2} - V_{1}) = 2 \times 1 0^{5} \times (3 - 1) \times 1 0^{- 3}$ . Yeh step kyun? Jab $P$ constant ho toh woh integral se bahar aa jaata hai: $W = \int P d V = P Δ V$ . PV curve ke neeche ka area ek rectangle hai.
Compute karo. $W = 2 \times 1 0^{5} \times 2 \times 1 0^{- 3} = 400$ J.
Internal energy. $Δ U = Q - W = 700 - 400 = + 300$ J.
(b) Isothermal. Ideal gas ke liye $U = U (T)$ ; $T$ constant ⇒ $Δ U = 0$ ⇒ $0 = Q - W \Rightarrow W = Q = 400$ J. Yeh step kyun? Isothermal woh special case hai jahan First Law force karta hai ki saari absorbed heat kaam ban jaaye.

Verify (a): $Pa \times m^{3} = \frac{N}{m ^{2}} \cdot m^{3} = N\cdotm = J$ , toh $W = 400$ J dimensionally sahi hai; $700 - 400 = 300$ J. ✓ Verify (b): $Δ U = Q - W = 400 - 400 = 0$ , constant $T$ ke saath consistent. ✓

Figure isobaric case ko ek rectangle ke roop mein dikhati hai jiska area kaam hai.

Word problem / exam twist — ek full cycle

Worked example Example 10 — Cyclic process (exam favourite)

Ek gas ko PV diagram par ek closed cycle mein le jaaya jaata hai: woh upar jaate waqt $500$ J heat absorb karti hai, wapas aate waqt $300$ J expel karti hai, aur loop mein net $200$ J kaam karti hai. (a) Poore cycle ke liye $Δ U$ kya hai? (b) Numbers ke saath consistency check karo.

Forecast: Ek cycle usi state $(P, V, T)$ par wapas aata hai. Kyunki $U$ ek state function hai, loop mein ghoomne par $Δ U = 0$ hona chahiye — path se koi farq nahi.

$Δ U = 0$ kyun? $U$ sirf state par depend karta hai; start state = end state ⇒ $Δ U_{cycle} = 0$ . Yeh step kyun? Yahi "state function" ka poora matlab hai — parent ke lake analogy se — same water level par wapas aa jaao.
Net heat. $Q_{net} = + 500 - 300 = + 200$ J.
Cycle par law apply karo. $Δ U = Q_{net} - W_{net} \Rightarrow 0 = 200 - W_{net} \Rightarrow W_{net} = 200$ J.

Verify: Stated net work $200$ J hai aur hamara law predict karta hai $200$ J — consistent. Kisi bhi cycle mein, net heat in = net work out ( $Q_{net} = W_{net}$ ), jo exactly Second law of thermodynamics ka seed hai (tum heat in se zyada work out nahi le sakte). ✓

Recall check

Recall Cell ko outcome se match karo

Adiabatic expansion — gas garam hoti hai ya thandi? ::: Thandi; $Q = 0$ toh $Δ U = - W < 0$ . Ideal gas ki free expansion — $Δ T$ kya hai? ::: Zero; $Q = 0$ , $W = 0$ ⇒ $Δ U = 0$ ⇒ $Δ T = 0$ . Isochoric process — kaun sa term gayab hota hai? ::: $W = 0$ , toh $Δ U = Q$ . Full cycle — $Δ U$ kya hai? ::: Zero, kyunki $U$ ek state function hai. Compression with heat loss (Ex 8) — $Δ U$ ka sign? ::: $+ 90$ J; compression ne heat remove se zyada add kiya.

Connections

Parent: First Law & sign conventions
Work done in thermodynamic processes (PV diagrams)
Isothermal, adiabatic, isobaric, isochoric processes
Internal energy and degrees of freedom
Heat capacities Cp and Cv
Conservation of energy (mechanics)
Second law of thermodynamics